POJ - 2559 Largest Rectangle in a Histogram【单调栈】

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Time limit 1000 ms
Memory limit 65536 kB

A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles:

Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.

Input

The input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that 1<=n<=100000. Then follow n integers h1,…,hn, where 0<=hi<=1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case.

Output

For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.

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题目分析

可以分别尝试以每个矩形的高作为答案矩形的高
当以 $h_i$为答案矩形的高时
分别找到第 $i$个矩形左边和右边第一个高度小于 $h_i$的矩形lh和rh
算出 $lh$到 $rh$之间的总宽度尝试是否能更新答案
而求左边/右边第一个高度小于 $hi$的矩形，显然可以用单调栈

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#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long lt;
 
lt read()
{
    lt f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

const int maxn=100010;
int n;
lt hi[maxn],ans;
int lh[maxn],rh[maxn];
int st[maxn],top;

int main()
{
   	while(scanf("%d",&n)!=EOF)
   	{
   		if(n==0) break; ans=top=0;
		for(int i=1;i<=n;++i) hi[i]=read();
   		for(int i=n;i>=1;--i)
   		{
   			while(top&&hi[i]<=hi[st[top]]) top--;
   			rh[i]=st[top]==0?n+1:st[top];
   			st[++top]=i;
		}
		top=0;
		for(int i=1;i<=n;++i)
		{
			while(top&&hi[i]<=hi[st[top]]) top--;
   			lh[i]=st[top];
   			st[++top]=i;
		}
		for(int i=1;i<=n;++i)
		ans=max(ans,hi[i]*(i-lh[i]+rh[i]-i-1));
		printf("%lld\n",ans);
	}
	return 0;
}